The Boston Diaries

The ongoing saga of a programmer who doesn't live in Boston, nor does he even like Boston, but yet named his weblog/journal “The Boston Diaries.”

Go figure.

Sunday, August 04, 2013

Programs from the past, Part II

I could have sworn I wrote a bit more about a program I wrote while at FAU; specifically, just how long it took to run back then vs now (or rather, a few years ago), but having failed to find any using Google or locally searching the entries on the server, I ended up spending all day yesterday reading through ten years of archives and still not finding anything, I guess I haven't.

Well. [Deep subject. —Editor] [Shut up you! —Sean]

So I wrote a series of programs to search through the domain of the following pair of equations:

xi+1 = (Ayi + B) xi (1 - xi)
yi+1 = (Cxi + D) yi (1 - yi)

The intent was to generate a large number of (x y) pairs and plot the results for a given set of values for A, B, C and D. Something like:

[Window showing a chaotic attractor]
[Control window for chaotic attractor]

The top window shows the resulting chaotic attractor when A is 2.4375, B is 1.5624, C is -0.8659 and D is 4.0 as you loop through the above two equations 15,000 times to generate 15,000 points (all between 0 and 1 for both the x and y axis). And the thing about a chaotic attractor is, the initial starting point (x0 and y0) is largely immaterial (but not quite, as the starting values must be between somewhere between 0 and 1). Any given value for x0 and y0 will result in the same figure being drawn for the given values of A, B, C and D.

The program shown allows you to change the settings of the four controlling variables A, B, C and D by sliding the named boxes horizontally (the botton one allows you to slide multiple variables at the same time—in this case, A and C will slide to the left if you slide the bottom box to the left).

That was fine, but my bosses at the time wanted another view generated—a “map” of the chaotic attractors, if you will. To generate this “map,” first you calculate how many points are generated before you settle into an attractor:

int mainloop(const double A,const double B,const double C,const double D)
{
  double xn,xn1;
  double yn,yn1;
  int    ix;
  int    iy;
  int    count;
  bool   pix[DIM][DIM];	/* have we seen this (xy) pair yet? */
  
  memset(pix,0,sizeof(pix));
  
  for (count = 0 , xn = yn = 0.5 ; count < MAXLOOP ; count++)
  {
    xn1 = ((A * yn) + B) * xn * (1.0 - xn);
    yn1 = ((C * xn) + D) * yn * (1.0 - yn);    
    xn  = xn1;
    yn  = yn1;
    
    /* exit if we're outside the bounds */
    
    if (xn <  0.0) return MAXLOOP;
    if (xn >= 1.0) return MAXLOOP;
    if (yn <  0.0) return MAXLOOP;
    if (yn >= 1.0) return MAXLOOP;
    
    ix  = (int)(xn * DIM);
    iy  = (int)(yn * DIM);
    
    if (pix[ix][iy])
      break;
    pix[ix][iy] = true;
  }
  
  return(count);
}

The assumption—if this returns MAXLOOP, there is no attractor for the given set of values for A, B, C and D; othersise it's an indication of how quickly (or how “showey”) we settle into the pattern.

Then it's a matter of going through a range of values and for the map images, since they're two dimensional, I only sweep through two of the four governing variables over their range:

for (A = -4.0 ; A <= 4.0 ; A += 0.016)
  for (B = -4.0 ; B <= 4.0 ; B += 0.016)
    plot(A,B,count_as_color(mainloop(A,B,-0.8659,4.0));

(count_as_color() just maps a count to a color)

And we end up with an image something like:

[Map of chaotic attractors]

In this image, A is the horizontal axis, going left to right from -4.0 to 4.0, and B is the veritcal axis, going, bottom to top, from -4.0 to 4.0; C and D are fixed, at -0.8659 and 4.0 respectively. The red plus in the upper right hand quadrant is the location of the “French horn” attractor above. The black area represents areas that have no chaotic attractor.

But my bosses didn't just want one map—no, they wanted a larger set. So, I generated a series of images as this, each image having a different value for C. And while the map may not change drastically—for instance, if we bump C up a notch to -0.8499:

[It's different; it's subtle, but it's different]

The resulting chaotic attractor is completely different:

[Window showing a slightly different chaotic attractor]
[Control window for the slightly different chaotic attractor]

And I generated a ton of maps by looping through values of C from -4.0 to 4.0—500 images in total.

Now, when I did this the first time, twenty years (or a bit more—maybe up to twenty-two years ago) I was doing this on a state of the art Unix workstation, an SGI workstation (perhaps you've seen their commerical?). It cost $30,000 and it took a year to generate the 500 images. Each individual image took some ten hours to generate.

Fast forward to today. On a laptop that is maybe two years old now. I reran the code and was able to generate an entire image (in fact, both of the map images above) in only 8 seconds.

With four cores, that means, on a two year old laptop that might have cost $1,500 (tops) would take 17 minutes what it took me a year to generate twenty years ago.

Oh, and about those eight seconds

Update on August 5th, 2013

Oops, I made a slight mistake

Update on Tuesday, August 6th, 2013

Mistakes were made alright.


About those eight seconds

In my previous post, I mentioned a program that took a year to run twenty-some years ago on a then state-of-the-art workstation could now be done in 17 minutes on a two year old laptop.

What I didn't mention is that the program twenty-some years ago was written in C and the program I ran today was written in Lua.

Yes, computers are so fast these days that a scripting language can out-perform computers from two decades ago.

“Okay,” you say. “But I won't want to wait 17 minutes for my data.”

Okay, fine. I see two options, and let's try the first option and one that most people would do—drop down to C. And yes, that does give us an improvement, an impressive improvement—only 2.5 seconds per frame, and across four cores that means you'll have the results in a little over five minutes. Not that bad.

The other option, and hear me out—is to take our Lua code and run it via LuaJIT, a (pretty much) drop in replacement for Lua that compiles down to native code. Even if it's a bit slower than C, it should still be faster than Lua with no code changes.

So how does LuaJIT fare?

Personally, I was expecting the C version (which I actually wrote first) to be faster, if only buy a little bit, but …

So, here's the C version (which generates a single image):

[spc]saltmine:~/source/play>time ./a.out >/dev/null

real    0m2.483s
user    0m2.470s
sys     0m0.000s

And now the LuaJIT version:

[spc]saltmine:~/source/play>time luajit amap.lua >/dev/null

real    0m0.849s
user    0m0.840s
sys     0m0.000s
[Yeah, that's what I did when I saw these results]

And no, that's not a mistake (belive me, I checked and rechecked)—here:

[spc]saltmine:~/source/play>time lua amap.lua >/dev/null

real    0m8.091s
user    0m8.060s
sys     0m0.000s
[spc]saltmine:~/source/play>time luajit amap.lua >/dev/null

real    0m0.856s
user    0m0.850s
sys     0m0.000s

So … um … I can have that data to you in two minutes? Is that fast enough?

On reflection, it makes sense that LuaJIT will outperform C in this case. It's heavily CPU bound and the fact that the main function:

function mainloop(A,B,C,D)
  local pix = {}
  local xn  = 0.5
  local yn  = 0.5
  
  for count = 1 , MAX do
    local xn1 = ((A * yn) + B) * xn * (1.0 - xn)
    local yn1 = ((C * xn) + D) * yn * (1.0 - yn)
    
    xn = xn1
    yn = yn1
    
    if xn <  0 then return MAX-1 end
    if xn >= 1 then return MAX-1 end
    if yn <  0 then return MAX-1 end
    if yn >= 1 then return MAX-1 end
    
    local ix = math.floor(xn * DIM)
    local iy = math.floor(yn * DIM)
    local f  = iy * DIM + ix -- Lua doesn't really do N-dimensional arrays
    
    if pix[f] then return count-1 end
    pix[f] = true
  end
end

can be recompiled per call to take advantage of the paramters. For instance, when A and B are both 0, the first expression then becomes:

local xn1 = xn * (1.0 - xn)

and given that I'm doing 32,768 interations of this (oh, did I fail to mention that? Yes, I'm doing 27,768 more interations than the code did twenty-some years ago) this does save quite a bit of time.

Update on August 5th, 2013

Oops, I made a slight mistake

Update on Tuesday, August 6th, 2013

Mistakes were made alright.

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